ism_chapter_03

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123 
Chapter 3 
Motion in Two and Three Dimensions 
 
Conceptual Problems 
 
*1 • 
Determine the Concept The distance traveled along a path can be represented as a 
sequence of displacements. 
 
Suppose we take a trip along some path and consider the trip as a sequence of many very 
small displacements. The net displacement is the vector sum of the very small 
displacements, and the total distance traveled is the sum of the magnitudes of the very 
small displacements. That is, 
 
total distance = NN 1,3,22,11,0 ... −∆++∆+∆+∆ rrrr rrrr 
 
where N is the number of very small displacements. (For this to be exactly true we have 
to take the limit as N goes to infinity and each displacement magnitude goes to zero.) 
Now, using ″the shortest distance between two points is a straight line,″ we have 
 
NNN 1,3,22,11,0,0 ... −∆++∆+∆+∆≤∆ rrrrr rrrrr , 
 
 where N,0r
r∆ is the magnitude of the net displacement. 
 
Hence, we have shown that the magnitude of the displacement of a particle is less than or 
equal to the distance it travels along its path. 
 
2 • 
Determine the Concept The displacement of an object is its final position vector minus 
its initial position vector ( if rrr
rrr −=∆ ). The displacement can be less but never more 
than the distance traveled. Suppose the path is one complete trip around the earth at the 
equator. Then, the displacement is 0 but the distance traveled is 2πRe. 
 
Chapter 3 
 
 
124 
3 • 
Determine the Concept The important distinction here is that average velocity is being 
requested, as opposed to average speed. 
 
The average velocity is defined as 
the displacement divided by the 
elapsed time. 
 
00av =∆=∆
∆=
tt
rv
rr
 
circuit. completeeach of end the
at zero always is velocity average thes,car travel race fast the howmatter 
no that see weThus zero. is track thearound any tripfor nt displaceme The
 
 
What is the correct answer if we were asked for average speed? 
 
The average speed is defined as the 
distance traveled divided by the 
elapsed time. 
 
t
v ∆≡
distancetotal
av 
zero.not is average theand zeroan greater th
be will traveleddistance total theany track, ofcircuit complete oneFor 
 
 
4 • 
False. Vectors are quantities with magnitude and direction that can be added and 
subtracted like displacements. Consider two vectors that are equal in magnitude and 
oppositely directed. Their sum is zero, showing by counterexample that the statement is 
false. 
 
5 • 
Determine the Concept We can answer 
this question by expressing the relationship 
between the magnitude of vector A
r
 and its 
component AS and then using properties of 
the cosine function. 
 
 
 
Express AS in terms of A and θ : AS = A cosθ 
 
Take the absolute value of both 
sides of this expression: 
 
⎜AS⎜ = ⎜A cosθ ⎜ = A⎜cosθ ⎜ 
and 
⎜cosθ ⎜=
A
AS 
Motion in One and Two Dimensions 
 
 
125
Using the fact that 0 < ⎟cosθ ⎜≤ 1, 
substitute for⎟cosθ ⎜to obtain: 10 S ≤< A
A
or AA ≤< S0 
 vector. theof magnitude theto
 equalor than less bemust vector a ofcomponent a of magnitude The No.
 
 
 equal. arecomponent its and vector theof magnitude the
 then,180 of multiplesor 0 toequal is figure in theshown angle theIf °°θ
 
 
*6 • 
Determine the Concept The diagram 
shows a vector A
r
and its components Ax 
and Ay. We can relate the magnitude of 
A
r
is related to the lengths of its 
components through the Pythagorean 
theorem. 
 
 
Suppose that A
r
is equal to zero. Then .0222 =+= yx AAA 
But .0022 ==⇒=+ yxyx AAAA 
 too.zero bemust components its ofeach zero, toequal is vector a If No. 
 
7 • 
Determine the Concept No. Consider the special case in which AB
rr −= . 
If 0then,0 =≠−= CAB rrr and the magnitudes of the components of BA rr and are 
larger than the components of .C
r
 
 
*8 • 
Determine the Concept The instantaneous acceleration is the limiting value, as ∆t 
approaches zero, of .t∆∆vr Thus, the acceleration vector is in the same direction as .vr∆ 
 
 
False. Consider a ball that has been 
thrown upward near the surface of 
the earth and is slowing down. The 
direction of its motion is upward. 
 
The diagram shows the ball’s 
velocity vectors at two instants of 
time and the determination of .vr∆ 
Note that because vr∆ is downward 
so is the acceleration of the ball. 
 
 
Chapter 3 
 
 
126 
9 • 
Determine the Concept The instantaneous acceleration is the limiting value, as ∆t 
approaches zero, of t∆∆vr and is in the same direction as .vr∆ 
 
Other than through the definition of ,ar the instantaneous velocity and acceleration vectors 
are unrelated. Knowing the direction of the velocity at one instant tells one nothing about 
how the velocity is changing at that instant. correct. is )(e 
 
10 • 
Determine the Concept The changing velocity of the golf ball during its flight can be 
understood by recognizing that it has both horizontal and vertical components. The nature 
of its acceleration near the highest point of its flight can be understood by analyzing the 
vertical components of its velocity on either side of this point. 
 
At the highest point of its flight, the 
ball is still traveling horizontally 
even though its vertical velocity is 
momentarily zero. The figure to the 
right shows the vertical components 
of the ball’s velocity just before and 
just after it has reached its highest 
point. The change in velocity during 
this short interval is a non-zero, 
downward-pointing vector. Because 
the acceleration is proportional to 
the change in velocity, it must also 
be nonzero. 
 
 
correct. is )(d 
Remarks: Note that vx is nonzero and vy is zero, while ax is zero and ay is nonzero. 
 
11 • 
Determine the Concept The change in the velocity is in the same direction as the 
acceleration. Choose an x-y coordinate system with east being the positive x direction 
and north the positive y direction. 
 
Given our choice of coordinate system, the x component of ar is negative and so vr will 
decrease. The y component of ar is positive and so vr will increase toward the north. 
correct. is )(c 
 
*12 • 
Determine the Concept The average velocity of a particle, ,avv
r
is the ratio of the 
particle’s displacement to the time required for the displacement. 
 
(a) We can calculate rr∆ from the given information and ∆t is known. correct. is )(a 
 
(b) We do not have enough information to calculate vr∆ and cannot compute the 
Motion in One and Two Dimensions 
 
 
127
particle’s average acceleration. 
 
(c) We would need to know how the particle’s velocity varies with time in order to 
compute its instantaneous velocity. 
 
(d) We would need to know how the particle’s velocity varies with time in order to 
compute its instantaneous acceleration. 
 
13 •• 
Determine the Concept The velocity vector is always in the direction of motion and, 
thus, tangent to the path. 
 
(a) 
path. theo tangent tis motion,
ofdirection in the being always of econsequenc a as vector, velocity The
 
 
(b) A sketch showing two velocity 
vectors for a particle moving along a 
path is shown to the right. 
 
 
14 • 
Determine the Concept An object experiences acceleration whenever either its speed 
changes or it changes direction. 
 
The acceleration of a car moving in a straight path at constant speed is zero. In the other 
examples, either the magnitude or the direction of the velocity vector is changing and, 
hence, the car is accelerated. correct. is )(b 
 
*15 • 
Determine the Concept The velocity vector is defined by ,/ dtdrv rr = while the 
acceleration vector is defined by ./ dtdva rr = 
 
(a) A car moving along a straight roadwhile braking. 
 
(b) A car moving along a straight road while speeding up. 
(c) A particle moving around a circular track at constant speed. 
 
16 • 
Determine the Concept A particle experiences accelerated motion when either its speed 
or direction of motion changes. 
 
A particle moving at constant speed in a circular path is accelerating because the 
Chapter 3 
 
 
128 
direction of its velocity vector is changing. 
 
If a particle is moving at constant velocity, it is not accelerating. 
 
17 •• 
Determine the Concept The acceleration vector is in the same direction as the change in 
velocity vector, .vr∆ 
 
(a) The sketch for the dart thrown 
upward is shown to the right. The 
acceleration vector is in the 
direction of the change in the 
velocity vector .vr∆ 
 
(b) The sketch for the falling dart is 
shown to the right. Again, the 
acceleration vector is in the 
direction of the change in the 
velocity vector .vr∆ 
 
 
 
(c) The acceleration vector is in the 
direction of the change in the 
velocity vector … and hence is 
downward as shown the right: 
 
 
 
*18 •• 
Determine the Concept The acceleration vector is in the same direction as the change in 
velocity vector, .vr∆ 
 
The drawing is shown to the right. 
 
 
19 •• 
Determine the Concept The acceleration vector is in the same direction as the change in 
velocity vector, .vr∆ 
 
The sketch is shown to the right. 
 
 
Motion in One and Two Dimensions 
 
 
129
20 • 
Determine the Concept We can decide what the pilot should do by considering the 
speeds of the boat and of the current. 
 
Give up. The speed of the stream is equal to the maximum speed of the boat in still 
water. The best the boat can do is, while facing directly upstream, maintain its position 
relative to the bank. correct. is )(d 
 
*21 • 
Determine the Concept True. In the absence of air resistance, both projectiles 
experience the same downward acceleration. Because both projectiles have initial vertical 
velocities of zero, their vertical motions must be identical. 
 
22 • 
Determine the Concept In the absence of air resistance, the horizontal component of the 
projectile’s velocity is constant for the duration of its flight. 
 
At the highest point, the speed is the horizontal component of the initial velocity. The 
vertical component is zero at the highest point. correct. is )(e 
 
23 • 
Determine the Concept In the absence of air resistance, the acceleration of the ball 
depends only on the change in its velocity and is independent of its velocity. 
 
As the ball moves along its trajectory between points A and C, the vertical component of 
its velocity decreases and the change in its velocity is a downward pointing vector. 
Between points C and E, the vertical component of its velocity increases and the change 
in its velocity is also a downward pointing vector. There is no change in the horizontal 
component of the velocity. correct. is )(d 
 
24 • 
Determine the Concept In the absence of air resistance, the horizontal component of the 
velocity remains constant throughout the flight. The vertical component has its maximum 
values at launch and impact. 
 
(a) The speed is greatest at A and E. 
 
(b) The speed is least at point C. 
 
(c) The speed is the same at A and E. The horizontal components are equal at these points 
but the vertical components are oppositely directed. 
 
25 • 
Determine the Concept Speed is a scalar quantity, whereas acceleration, equal to the 
rate of change of velocity, is a vector quantity. 
(a) False. Consider a ball on the end of a string. The ball can move with constant speed 
Chapter 3 
 
 
130 
(a scalar) even though its acceleration (a vector) is always changing direction. 
(b) True. From its definition, if the acceleration is zero, the velocity must be constant and 
so, therefore, must be the speed. 
 
26 • 
Determine the Concept The average acceleration vector is defined by ./av t∆∆= va rr 
 
The direction of ava
r
is that of 
,if vvv
rrr −=∆ as shown to the right. 
 
 
27 • 
Determine the Concept The velocity of B relative to A is .ABBA vvv
rrr −= 
 
The direction of ABBA vvv
rrr −= is shown to 
the right. 
 
 
*28 •• 
(a) The vectors ( )tAr and ( )tt ∆+Ar are of equal length but point in slightly different 
directions. A
r∆ is shown in the diagram below. Note that Ar∆ is nearly perpendicular 
to ( )tAr . For very small time intervals, Ar∆ and ( )tAr are perpendicular to one another. 
Therefore, dtd /A
r
is perpendicular to .A
r
 
 
(b) If A
r
represents the position of a particle, the particle must be undergoing circular 
motion (i.e., it is at a constant distance from some origin). The velocity vector is tangent 
to the particle’s trajectory; in the case of a circle, it is perpendicular to the circle’s radius. 
 
(c) Yes, it could in the case of uniform circular motion. The speed of the particle is 
constant, but its heading is changing constantly. The acceleration vector in this case is 
Motion in One and Two Dimensions 
 
 
131
always perpendicular to the velocity vector. 
 
29 •• 
Determine the Concept The velocity vector is in the same direction as the change in the 
position vector while the acceleration vector is in the same direction as the change in the 
velocity vector. Choose a coordinate system in which the y direction is north and the x 
direction is east. 
(a) 
Path Direction of velocity 
vector 
AB north 
BC northeast 
CD east 
DE southeast 
EF south 
(b) 
Path Direction of acceleration 
vector 
AB north 
BC southeast 
CD 0 
DE southwest 
EF north 
 
(c) 
ere.smaller th ispath 
 theof radius thesince DEfor larger but ,comparable are magnitudes The
 
 
*30 •• 
Determine the Concept We’ll assume that the cannons are identical and use a constant-
acceleration equation to express the displacement of each cannonball as a function of 
time. Having done so, we can then establish the condition under which they will have the 
same vertical position at a given time and, hence, collide. The modified diagram shown 
below shows the displacements of both cannonballs. 
 
 
Express the displacement of the 
cannonball from cannon A at any 
time t after being fired and before 
any collision: 
 
2
2
1
0 tt gvr
rrr +=∆ 
Express the displacement of the 
cannonball from cannon A at any 
time t′ after being fired and before 
any collision: 
 
 
2
2
1
0 tt ′+′′=′∆ gvr rrr 
Chapter 3 
 
 
132 
usly.simultaneo
guns thefire should they Therefore, times.allat sight of line thebelow 
distance same theare balls theand ' usly,simultaneo fired are guns theIf
2
2
1 gt
tt =
 
 
Remarks: This is the ″monkey and hunter″ problem in disguise. If you imagine a 
monkey in the position shown below, and the two guns are fired simultaneously, and 
the monkey begins to fall when the guns are fired, then the monkey and the two 
cannonballs will all reach point P at the same time. 
 
 
 
31 •• 
Determine the Concept The droplet leaving the bottle has the same horizontal velocity 
as the ship. During the time the droplet is in the air, it is also moving horizontally with 
the same velocity as the rest of the ship. Because of this, it falls into the vessel, which 
has the same horizontal velocity. Because you have the same horizontal velocity as the 
ship does, you see the same thing as if the ship were standing still. 
 
32 • 
Determine the Concept 
 
(a) 
stone. theof velocity therepresent 
 could themofeither stone, theofpath theo tangent tare and Because DA
rr
 
 
 
 
(b) 
( ) ( )
( ) ( )
stone. theofon accelerati therepresent could vector only the
 and lartoperpendicu is / Therefore, another. one lar toperpendicu
 are and intervals, timesmallFor very . lar toperpendicu
nearly is that Note above. diagram in theshown are and vectors
 twoThese circle. thearound moves stone theas directionsdifferent 
slightlyin point but length equal of be and vectorsLet the
E
AA
AAA
AA
BA
r
rr
rrr
rr
rr
dtd
tt
ttt
∆
∆∆
∆+
 
 
Motion in One and Two Dimensions 
 
 
133
33 • 
Determine the Concept True. An object accelerates when its velocity changes; that is, 
when either its speed or its direction changes. When an object moves in a circle the 
direction of its motion is continually changing. 
 
34 •• 
Picture the Problem In the diagram, (a) 
shows the pendulum just before it reverses 
direction and (b) shows the pendulum just 
after it has reversed its direction. The 
acceleration of the bob is in the direction of 
the change in the velocity if vvv
rrr −=∆ and 
is tangent to the pendulum trajectory at the 
point of reversal of direction. This makes 
sense because, at an extremum of motion, 
v = 0, so there is no centripetal 
acceleration. However, because the 
velocity is reversing direction, the 
tangential acceleration is nonzero. 
 
 
 
35 • 
Determine the Concept The principle reason is aerodynamic drag. When moving 
through a fluid, such as the atmosphere, the ball's acceleration will depend strongly on its 
velocity. 
 
Estimation and Approximation 
 
*36 •• 
Picture the Problem During the flight of the ball the acceleration is constant and equal 
to 9.81 m/s2 directed downward. We can find the flight time from the vertical part of the 
motion, and then use the horizontal part of the motion to find the horizontal distance. 
We’ll assume that the release point of the ball is 2 m above your feet. 
 
Make a sketch of the motion. 
Include coordinate axes, initial and 
final positions, and initial velocity 
components: 
 
Obviously, how far you throw the 
ball will depend on how fast you 
can throw it. A major league 
baseball pitcher can throw a fastball 
at 90 mi/h or so. Assume that you 
can throw a ball at two-thirds that 
speed to obtain: 
m/s8.26
mi/h1
m/s0.447mi/h600 =×=v 
 
Chapter 3 
 
 
134 
There is no acceleration in the x 
direction, so the horizontal motion is 
one of constant velocity. Express the 
horizontal position of the ball as a 
function of time: 
 
tvx x0= (1) 
 
Assuming that the release point of 
the ball is a distance h above the 
ground, express the vertical position 
of the ball as a function of time: 
 
2
2
1
0 tatvhy yy ++= (2) 
 
(a) For θ = 0 we have: 
 
( )
m/s8.26
0cosm/s8.26cos 000
=
°== θvv x 
and ( ) 00sinm/s8.26sin 000 =°== θvv y 
 
Substitute in equations (1) and (2) to 
obtain: 
 
( )tx m/s8.26= 
and ( ) 2221 m/s81.9m2 ty −+= 
 
Eliminate t between these equations 
to obtain: 
 ( )
2
2
2
m/s8.26
m/s4.91m2 xy −= 
 
At impact, y = 0 and x = R: 
 ( ) 22
2
m/s8.26
m/s4.91m20 R−= 
 
Solve for R to obtain: 
 
m1.17=R 
 
(b) Using trigonometry, solve for v0x 
and v0y: 
( )
m/s0.19
45cosm/s8.26cos 000
=
°== θvv x 
and ( )
m/s0.19
45sinm/s8.26sin 000
=
°== θvv y 
 
Substitute in equations (1) and (2) to 
obtain: 
 
( )tx m/s0.19= 
and ( ) ( ) 2221 m/s81.9m/s19.0m2 tty −++= 
 
Eliminate t between these equations 
to obtain: 
 ( )
2
2
2
m/s0.19
m/s4.905m2 xxy −+= 
 
Motion in One and Two Dimensions 
 
 
135
At impact, y = 0 and x = R. Hence: 
 ( ) 22
2
m/s0.19
m/s4.905m20 RR −+= 
or ( ) 0m2.147m60.73 22 =−− RR 
 
Solve for R (you can use the 
″solver″ or ″graph″ functions of 
your calculator) to obtain: 
 
m6.75=R 
(c) Solve for v0x and v0y: m/s8.2600 == vv x 
and 
00 =yv 
 
Substitute in equations (1) and (2) to 
obtain: 
 
( )tx m/s8.26= 
and ( ) 2221 m/s81.9m14 ty −+= 
 
Eliminate t between these equations 
to obtain: 
 ( )
2
2
2
m/s8.26
m/s4.905m14 xy −= 
 
At impact, y = 0 and x = R: 
 ( ) 22
2
m/s8.26
m/s4.905m140 R−= 
 
Solve for R to obtain: 
 
m3.45=R 
 
(d) Using trigonometry, solve for 
v0x and v0y: 
 
 v0x = v0 y = 19.0 m / s 
 
Substitute in equations (1) and (2) to 
obtain: 
 
( )tx m/s0.19= 
and ( ) ( ) 2221 m/s81.9m/s19.0m14 tty −++= 
 
Eliminate t between these equations 
to obtain: 
 ( )
2
2
2
m/s0.19
m/s4.905m14 xxy −+= 
 
At impact, y = 0 and x = R: 
 ( ) 22
2
m/s0.19
m/s4.905m140 RR −+= 
 
Solve for R (you can use the 
″solver″ or ″graph″ function of your 
calculator) to obtain: 
m6.85=R 
 
Chapter 3 
 
 
136 
37 •• 
Picture the Problem We’ll ignore the height of Geoff’s release point above the ground 
and assume that he launched the brick at an angle of 45°. Because the velocity of the 
brick at the highest point of its flight is equal to the horizontal component of its initial 
velocity, we can use constant-acceleration equations to relate this velocity to the brick’s x 
and y coordinates at impact. The diagram shows an appropriate coordinate system and the 
brick when it is at point P with coordinates (x, y). 
 
 
 
Using a constant-acceleration 
equation, express the x coordinate of 
the brick as a function of time: 
 
2
2
1
00 tatvxx xx ++= 
or, because x0 = 0 and ax = 0, 
tvx x0= 
 
Express the y coordinate of the brick 
as a function of time: 
 
2
2
1
00 tatvyy yy ++= 
or, because y0 = 0 and ay = −g, 
2
2
1
0 gttvy y −= 
 
Eliminate the parameter t to obtain: ( ) 22
0
0 2
tan x
v
gxy
x
−= θ 
 
Use the brick’s coordinates when it 
strikes the ground to obtain: ( ) 22
0
0 2
tan0 R
v
gR
x
−= θ 
where R is the range of the brick. 
Solve for v0x to obtain: 
0
0 tan2 θ
gRv x = 
 
Substitute numerical values and 
evaluate v0x: 
( )( ) m/s8.14
45tan2
m44.5m/s9.81 2
0 =°=xv 
Note that, at the brick’s highest point, 
vy = 0. 
 
Vectors, Vector Addition, and Coordinate Systems 
 
38 • 
Picture the Problem Let the positive y direction be straight up, the positive x direction 
be to the right, and A
r
and B
r
be the position vectors for the minute and hour hands. The 
Motion in One and Two Dimensions 
 
 
137
pictorial representation below shows the orientation of the hands of the clock for parts (a) 
through (d). 
 
 
 
(a) The position vector for the 
minute hand at12:00 is: 
 
( ) jA ˆm5.000:12 =r 
The position vector for the hour 
hand at 12:00 is: 
 
( ) jB ˆm25.000:12 =r 
(b) At 3:30, the minute hand is positioned along the −y axis, while the hour hand is at an 
angle of (3.5 h)/12 h × 360° = 105°, measured clockwise from the top. 
 
The position vector for the minute 
hand is: 
 
( ) jA ˆm5.030:3 −=r 
Find the x-component of the vector 
representing the hour hand: 
 
( ) m241.0105sinm25.0 =°=xB 
Find the y-component of the vector 
representing the hour hand: 
 
( ) m0647.0105cosm25.0 −=°=yB 
The position vector for the hour 
hand is: 
 
( ) ( ) jiB ˆm0647.0ˆm241.030:3 −=r 
(c) At 6:30, the minute hand is positioned along the −y axis, while the hour hand is at an 
angle of (6.5 h)/12 h × 360° = 195°, measured clockwise from the top. 
 
The position vector for the minute 
hand is: 
 
( ) jA ˆm5.030:6 −=r 
Find the x-component of the vector 
representing the hour hand: 
 
( ) m0647.0195sinm25.0 −=°=xB 
Find the y-component of the vector 
representing the hour hand: 
 
( ) m241.0195cosm25.0 −=°=yB 
The position vector for the hour 
hand is: 
 
( ) ( ) jiB ˆm241.0ˆm0647.030:6 −−=r 
(d) At 7:15, the minute hand ispositioned along the +x axis, while the hour hand is at an 
angle of (7.25 h)/12 h × 360° = 218°, measured clockwise from the top. 
 
Chapter 3 
 
 
138 
The position vector for the minute 
hand is: 
 
( )iA ˆm5.015:7 =r 
Find the x-component of the vector 
representing the hour hand: 
 
( ) m154.0218sinm25.0 −=°=xB 
Find the y-component of the vector 
representing the hour hand: 
 
( ) m197.0218cosm25.0 −=°=yB 
The position vector for the hour 
hand is: 
 
( ) ( ) jiB ˆm197.0ˆm154.015:7 −−=r 
 
(e) Find A
r
 − Br at 12:00: ( ) ( )
( ) j
jjBA
ˆm25.0
ˆm25.0ˆm5.0
=
−=− rr
 
 
Find A
r
 − Br at 3:30: ( )
( ) ( )[ ]
( ) ( ) ji
ji
jBA
ˆm435.0ˆm241.0
ˆm0647.0ˆm241.0
ˆm5.0
−−=
−−
−=− rr
 
 
Find A
r
 − Br at 6:30: ( )
( ) ( )[ ]
( ) ( ) ji
ji
jBA
ˆm259.0ˆm0647.0
ˆm241.0ˆm0647.0
ˆm5.0
−−=
−−
−=− rr
 
 
Find A
r
 − Br at 7:15: ( )
( ) ( )[ ]
( ) ( ) ji
ji
jBA
ˆm697.0ˆm152.0
ˆm197.0ˆm152.0
ˆm5.0
+=
−−−
=− rr
 
 
 
*39 • 
Picture the Problem The resultant displacement is the vector sum of the individual 
displacements. 
 
The two displacements of the bear 
and its resultant displacement are 
shown to the right: 
 
 
 
 
 
 
 
Motion in One and Two Dimensions 
 
 
139
Using the law of cosines, solve for 
the resultant displacement: 
 
( ) ( )
( )( ) °−
+=
cos135m12m122
m12m12 222R
 
and 
m2.22=R 
 
Using the law of sines, solve for α: 
m2.22
135sin
m12
sin °=α 
∴ α = 22.5° and the angle with the 
horizontal is 45° − 22.5° = °5.22 
 
40 • 
Picture the Problem The resultant displacement is the vector sum of the individual 
displacements. 
 
(a) Using the endpoint coordinates 
for her initial and final positions, 
draw the student’s initial and final 
position vectors and construct her 
displacement vector. 
 
 
 
 
 
Find the magnitude of her 
displacement and the angle this 
displacement makes with the 
positive x-axis: 
 
.135 @ m 25 isnt displacemeHer ° 
 
(b) 
.135 @ 25 also
isnt displaceme his so ),(in as same theare positions final and initial His
°
a
 
 
 
*41 • 
Picture the Problem Use the standard rules for vector addition. Remember that 
changing the sign of a vector reverses its direction. 
 
 (a) 
 
(b) 
 
Chapter 3 
 
 
140 
(c) 
 
 
(d) 
 
 
(e) 
 
 
 
42 • 
Picture the Problem The figure 
shows the paths walked by the 
Scout. The length of path A is 
2.4 km; the length of path B is 
2.4 km; and the length of path C is 
1.5 km: 
 
 
(a) Express the distance from the 
campsite to the end of path C: 
 
2.4 km – 1.5 km = km9.0 
(b) Determine the angle θ subtended 
by the arc at the origin (campsite): 
°==
==
57.3rad1
km2.4
km2.4
radius
lengtharc
radiansθ 
east.ofnorth rad 1 is
camp fromdirectionHis
 
 
(c) Express the total distance as the 
sum of the three parts of his walk: 
 
 
 
dtot = deast + darc + dtoward camp 
 
 
 
 
Substitute the given distances to 
find the total: 
 
dtot = 2.4 km + 2.4 km + 1.5 km 
 = 6.3 km 
 
Motion in One and Two Dimensions 
 
 
141
Express the ratio of the magnitude 
of his displacement to the total 
distance he walked and substitute to 
obtain a numerical value for this 
ratio: 7
1
km6.3
km0.9
 walkeddistance Total
ntdisplaceme his of Magnitude
=
=
 
 
43 • 
Picture the Problem The direction of a vector is determined by its components. 
 
°−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= − 5.32
m/s5.5
m/s3.5tan 1θ 
The vector is in the fourth quadrant and 
correct. is )(b 
 
44 • 
Picture the Problem The components of the resultant vector can be obtained from the 
components of the vectors being added. The magnitude of the resultant vector can then 
be found by using the Pythagorean Theorem. 
 
A table such as the one shown to the 
right is useful in organizing the 
information in this problem. Let D
r
 
be the sum of vectors ,A
r
,B
r
 and .C
r
 
 
 
 
 
Vector x-component y-component 
A
r
 6 −3 
B
r
 −3 4 
C
r
 2 5 
D
r
 
 
Determine the components of D
r
by 
adding the components of ,A
r
,B
r
 
and .C
r
 
 
Dx = 5 and Dy = 6 
 
Use the Pythagorean Theorem to 
calculate the magnitude of D
r
: 
( ) ( ) 81.765 2222 =+=+= yx DDD 
and correct. is )(d 
 
45 • 
Picture the Problem The components of the given vector can be determined using right-
triangle trigonometry. 
 
Use the trigonometric relationships between the magnitude of a vector and its 
components to calculate the x- and y-components of each vector. 
 A θ Ax Ay 
(a) 10 m 30° 8.66 m 5 m 
(b) 5 m 45° 3.54 m 3.54 m 
(c) 7 km 60° 3.50 km 6.06 km 
(d) 5 km 90° 0 5 km 
Chapter 3 
 
 
142 
(e) 15 km/s 150° −13.0 km/s 7.50 km/s 
(f) 10 m/s 240° −5.00 m/s −8.66 m/s 
(g) 8 m/s2 270° 0 −8.00 m/s2
 
*46 • 
Picture the Problem Vectors can be added and subtracted by adding and subtracting 
their components. 
 
Write A
r
in component form: 
 
Ax = (8 m) cos 37° = 6.4 m 
Ay = (8 m) sin 37° = 4.8 m 
∴ ( ) ( ) jiA ˆm8.4ˆm4.6 +=r 
 
(a), (b), (c) Add (or subtract) x- and 
y-components: 
 
( ) ( )
( ) ( )
( ) ( ) jiF
jiE
jiD
ˆm8.23ˆm6.17
ˆm8.9ˆm4.3
ˆm8.7ˆm4.0
+−=
−−=
+=
r
r
r
 
 
(d) Solve for G
r
and add 
components to obtain: 
( )
( ) ( ) ji
CBAG
ˆm9.2ˆm3.1
2
2
1
−=
++−= rrrr
 
 
47 •• 
Picture the Problem The magnitude of each vector can be found from the Pythagorean 
theorem and their directions found using the inverse tangent function. 
 
(a) jiA ˆ3ˆ5 +=r 
 
 
 
83.522 =+= yx AAA 
and, because A
r
is in the 1st quadrant, 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 0.31tan 1
x
y
A
Aθ 
 
(b) jiB ˆ7ˆ10 −=r 
 
 
2.1222 =+= yx BBB 
and, because B
r
is in the 4th quadrant, 
°−=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 0.35tan 1
x
y
B
Bθ 
 
(c) kjiC ˆ4ˆ3ˆ2 +−−=r 39.5222 =++= zyx CCCC 
°=⎟⎠
⎞⎜⎝
⎛= − 1.42cos 1
C
Czθ 
where θ is the polar angle measured from 
the positive z-axis and 
Motion in One and Two Dimensions 
 
 
143
 
 
°=⎟⎠
⎞⎜⎝
⎛ −=⎟⎠
⎞⎜⎝
⎛= −− 112
29
2coscos 11
C
Cxφ 
 
 
48 • 
Picture the Problem The magnitude and direction of a two-dimensional vector can be 
found by using the Pythagorean Theorem and the definition of the tangent function. 
 
 (a) jiA ˆ7ˆ4 −−=r 
 
 
 
 
 06.822 =+= yx AAA 
and, because A
r
is in the 3rd quadrant, 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 240tan 1
x
y
A
Aθ 
 
jiB ˆ2ˆ3 −=r 61.322 =+= yx BBB 
and, because B
r
is in the 4th quadrant, 
°−=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 7.33tan 1
x
y
B
Bθ 
 
jiBAC ˆ9ˆ −−=+= rrr 06.922 =+= yx CCC 
and, because C
r
is in the 3rd quadrant, 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 264tan 1
x
y
C
Cθ 
 
(b) Follow the same steps as in (a). A = 12.4 ; θ = °− 0.76 
B = 6.32 ; θ = °71.6 
C = 3.61 ; θ = °33.7 
 
49 • 
Picture the Problem The components of these vectors are related to the magnitude of 
each vector through the Pythagorean Theorem and trigonometric functions. In parts (a) 
and (b), calculate the rectangular components of each vector and then express the vector 
in rectangular form. 
 
Chapter 3 
 
 
144 
(a) Express vr in rectangular form: jiv ˆˆ yx vv +=r 
 
Evaluate vx and vy: vx = (10 m/s) cos 60° = 5 m/s 
and 
vy = (10 m/s) sin 60° = 8.66 m/s 
 
Substitute to obtain: jiv ˆ)m/s66.8(ˆ)m/s5( +=r 
 
(b) Express vr in rectangular form: 
 
jiA ˆˆ yx AA +=
r
 
Evaluate Ax and Ay: 
 
Ax = (5 m) cos 225° = −3.54 m 
and 
Ay = (5 m) sin 225°= −3.54 m 
 
Substitute to obtain: ( ) ( ) jiA ˆm54.3ˆm54.3 −+−=r 
(c) There is nothing to calculate as 
we are given the rectangular 
components: 
( ) ( ) jir ˆm6ˆm14 −=r 
 
50 • 
Picture the Problem While there are infinitely many vectors B that can be constructed 
such that A = B, the simplest are those which lie along the coordinate axes. 
 
Determine the magnitude of :A
r
 
 
543 2222 =+=+= yx AAA 
 
Write three vectors of the same 
magnitude as :A
r
 
 
jBiBiB ˆ5andˆ5ˆ5 321 =−==
rrr
,, 
 
The vectors are shown to the right: 
 
 
Motion in One and Two Dimensions 
 
 
145
*51 •• 
Picture the Problem While there are 
several walking routes the fly could take to 
get from the origin to point C, its 
displacement will be the same for all of 
them. One possible route is shown in the 
figure. 
 
 
Express the fly’s 
displacement D
r
during its trip from 
the origin to point C and find its 
magnitude: 
( ) ( ) ( )kji
CBAD
ˆm3ˆm3ˆm3 ++=
++= rrrr
 
and 
( ) ( ) ( )
m20.5
m3m3m3 222
=
++=D
 
 
*52 • 
Picture the Problem The diagram shows the locations of the transmitters relative to the 
ship and defines the distances separating the transmitters from each other and from the 
ship. We can find the distance between the ship and transmitter B using trigonometry. 
 
 
 
Relate the distance between A and B 
to the distance from the ship to A 
and the angle θ: 
 
SB
ABtan
D
D=θ 
Solve for and evaluate the distance 
from the ship to transmitter B: 
 
km173
30tan
km100
tan
AB
SB =°== θ
DD 
 
Chapter 3 
 
 
146 
Velocity and Acceleration Vectors 
 
53 • 
Picture the Problem For constant speed 
and direction, the instantaneous velocity is 
identical to the average velocity. Take the 
origin to be the location of the stationary 
radar and construct a pictorial 
representation. 
 
 
Express the average velocity: 
 t
r
∆
∆=
rr
avv 
Determine the position vectors: 
 
( )
( ) ( ) jir
jr
ˆkm1.14ˆkm1.14
and
ˆkm10
2
1
−+=
−=
r
r
 
 
Find the displacement vector: 
 ( ) ( ) ji
rrr
ˆkm1.4ˆkm1.14
12
−+=
−=∆ rrr
 
 
Substitute for rr∆ and ∆t to find the 
average velocity. 
( ) ( )
( ) ( ) ji
jiv
ˆkm/h1.4ˆkm/h1.14
h1
ˆkm1.4ˆkm1.14
av
−+=
−+=r
 
 
54 • 
Picture the Problem The average velocity is the change in position divided by the 
elapsed time. 
 
(a) The average velocity is: 
 
 t
rv ∆
∆=av 
Find the position vectors and the 
displacement vector: 
 
( ) ( ) jir ˆm3ˆm20 +=r 
( ) ( ) jir ˆm7ˆm62 +=r 
and 
( ) ( ) jirrr ˆm4ˆm412 +=−=∆ rrr 
 
Find the magnitude of the 
displacement vector for the interval 
between t = 0 and t = 2 s: 
 
( ) ( ) m66.5m4m4 2202 =+=∆r 
Motion in One and Two Dimensions 
 
 
147
Substitute to determine vav: 
 m/s83.2s2
m66.5
av ==v 
and 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 0.45
m4
m4tan 1θ measured 
from the positive x axis. 
 
(b) Repeat (a), this time using the 
displacement between t = 0 and 
t = 5 s to obtain: 
( ) ( ) jir ˆm14ˆm135 +=r , 
( ) ( ) jirrr ˆm11ˆm110505 +=−=∆ rrr , 
( ) ( ) m15.6m11m11 2205 =+=∆r , 
m/s3.11
s5
m15.6
av ==v , 
and 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 0.45
m11
m11tan 1θ measured 
from the positive x axis. 
 
*55 • 
Picture the Problem The magnitude of the velocity vector at the end of the 2 s of 
acceleration will give us its speed at that instant. This is a constant-acceleration problem. 
 
Find the final velocity vector of the 
particle: ( ) ( )( )
( ) ( ) ji
ji
jijiv
ˆm/s0.6ˆm/s0.4
ˆs0.2m/s0.3ˆm/s0.4
ˆˆˆˆ
2
0
+=
+=
+=+= tavvv yxyxr
 
Find the magnitude of :vr ( ) ( ) m/s7.21m/s6.0m/s4.0 22 =+=v 
and correct. is )(b 
 
56 • 
Picture the Problem Choose a coordinate system in which north coincides with the 
positive y direction and east with the positive x direction. Expressing the west and north 
velocity vectors is the first step in determining vr∆ and avar . 
 
(a) The magnitudes of 
NW and vv
rr
are 40 m/s and 30 m/s, 
respectively. The change in the 
magnitude of the particle’s velocity 
during this time is: 
 
m/s10
WN
−=
−=∆ vvv
 
Chapter 3 
 
 
148 
(b) The change in the direction of 
the velocity is from west to north. 
The change in direction is °90 
 
(c) The change in velocity is: 
 
 
( ) ( )
( ) ( ) ji
ijvvv
ˆm/s30ˆm/s40
ˆm/s40ˆm/s30WN
+=
−−=−=∆ rrr 
 
Calculate the magnitude and 
direction of :vr∆ ( ) ( ) m/s50m/s30m/s40 22 =+=∆v 
and 
°== −+ 9.36m/s40
m/s30tan 1axisxθ 
 
(d) Find the average acceleration 
during this interval: 
 
 
 
( ) ( )
( ) ( )ji
jiva
ˆm/s6ˆm/s8
s5
ˆm/s30ˆm/s40
22
av
+=
+=∆∆≡ trr 
The magnitude of this vector is: 
 ( ) ( ) 22222av m/s10m/s6m/s8 =+=a 
and its direction is 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 9.36
m/s8
m/s6tan 2
2
1θ measured 
from the positive x axis. 
 
57 • 
Picture the Problem The initial and final positions and velocities of the particle are 
given. We can find the average velocity and average acceleration using their definitions 
by first calculating the given displacement and velocities using unit vectors .ˆ and ˆ ji 
 
(a) The average velocity is: t∆∆≡ rv rrav 
 
The displacement of the particle 
during this interval of time is: 
 
( ) ( ) jir ˆm80ˆm100 +=∆r 
Substitute to find the average 
velocity: 
 
( ) ( )
( ) ( ) ji
jiv
ˆm/s7.26ˆm/s3.33
s3
ˆm80ˆm100
av
+=
+=r
 
 
(b) The average acceleration is: 
 
t∆∆= va rrav 
 
Motion in One and Two Dimensions 
 
 
149
Find vvv vrr ∆and,, 21 : 
 
( ) ( )
( ) ( )
( ) ( ) jiv
jiv
jiv
ˆm/s30.5ˆm/s00.9
ˆm/s0.23ˆm/s3.19
and
ˆm/s3.28ˆm/s3.28
2
1
−+−=∆∴
+=
+=
r
r
r
 
 
Using ∆t = 3 s, find the average 
acceleration: ( ) ( )jia ˆm/s77.1ˆm/s00.3 22av −+−=r 
 
*58 •• 
Picture the Problem The acceleration is constant so we can use the constant-acceleration 
equations in vector form to find the velocity at t = 2 s and the position vector at t = 4 s. 
 
(a) The velocity of the particle, as a 
function of time, is given by: 
 
tavv rrr += 0 
 
Substitute to find the velocity at 
t = 2 s: 
 [ ]( )
ji
ji
jiv
ˆm/s) 3( ˆm/s) (10
s2ˆ)m/s (3ˆ)m/s (4
ˆm/s) 9(ˆm/s) (2
22
−+=
++
−+=r
 
 
(b) Express the position vector as a 
function of time: 
 
2
2
1
00 tt avrr
rrrr ++= 
Substitute and simplify: 
 [ ]( )[ ]( )
ji
ji
ji
jir
ˆm) 9( ˆm) (44
s4ˆ)m/s (3 ˆ)m/s (4
s4ˆm/s) (-9 ˆm/s) (2
ˆm) (3 ˆm) (4 
222
2
1
−+=
++
++
+=r
 
 
Find the magnitude and direction of 
rr at t = 4 s: ( ) ( ) m9.44m9m44s) (4 22 =−+=r 
and, because rr is in the 4th quadrant, 
°−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= − 6.11
m44
m9tan 1θ 
 
59 •• 
Picture the Problem The velocity vector is the time-derivative of the position vector and 
the acceleration vector is the time-derivative of the velocity vector. 
 
Differentiate rr with respect to time: ( ) ( )[ ]
( ) ji
jirv
ˆ1040ˆ30
ˆ540ˆ30 2
t
ttt
dt
d
dt
d
−+=
−+==
rr
 
Chapter 3 
 
 
150 
where vr has units of m/s if t is in seconds. 
Differentiate vr with respect to time: ( )[ ]
( )j
jia
ˆm/s10
ˆ1040ˆ30
2−=
−+== t
dt
d
dt
vdrr
 
 
60 •• 
Picture the Problem We can use the constant-acceleration equations in vector form to 
solve the first part of the problem. In the second part, we can eliminate the parameter t 
from the constant-acceleration equations and express y as a function of x. 
 
 
(a) Use 0with 00 =+= vavv rrrr t to 
find :vr 
 
( ) ( )[ ]tjiv ˆm/s4ˆm/s622 +=r 
Use 22
1
00 tt avrr
rrrr ++= with ( )ir ˆm100 =r to find :rr 
 
( ) ( )[ ] ( )[ ]jir ˆm/s2ˆm/s3m10 2222 tt ++=r 
 
(b) Obtain the x and y components 
of the path from the vector equation 
in (a): 
( ) 22m/s3m10 tx += 
and ( ) 22m/s2 ty = 
Eliminate the parameter t from these 
equations and solve for y to obtain: 
 
 m
3
20 
3
2 −= xy 
Use this equation to plot the graph shown below. Note that the path in the xy plane is a 
straight line. 
 
0
2
4
6
8
10
12
14
16
18
20
0 10 20 30 40
x (m)
y 
(m
)
 
Motion in One and Two Dimensions 
 
 
151
61 ••• 
Picture the Problem The displacements 
of the boat are shown in the figure. We 
need to determine each of the 
displacements in order to calculate the 
average velocity of the boat during the 30-
s trip. 
 
 
(a) Express the average velocity of 
the boat: 
 
 
t∆
∆= rv
rr
av 
 
Express its total displacement: 
 ( ) ( )ij
rrr
ˆˆ
 
WW
2
NN2
1
WN
−∆+∆=
∆+∆=∆
tvta
rrr
 
 
 
To calculate the displacement we 
first have to find the speed after the 
first 20 s: 
 
 
 vW = vN, f = aN∆tN = 60 m/s 
so 
( ) ( )
( ) ( )ij
ijr
ˆm600 ˆm600 
ˆm/s 60ˆ W
2
NN2
1
−=
∆−∆=∆ ttar
 
 
Substitute to find the average 
velocity: 
( )( )
( )( )ji
jirv
ˆˆm/s20
s30
ˆˆm600
av
+−=
+−=∆
∆=
t
rr
 
(b) The average acceleration is 
given by: 
 ( ) ( )ii
vvra
ˆm/s2
s30
0ˆm/s60 2
if
av
−=−−=
∆
−=∆
∆=
tt
rrrr
 
 
(c) The displacement of the boat 
from the dock at the end of the 30-s 
trip was one of the intermediate 
results we obtained in part (a). 
( ) ( )
( )( )ji
ijr
ˆˆm600
ˆm600ˆm600
+−=
−+=∆r
 
 
Chapter 3 
 
 
152 
*62 ••• 
Picture the Problem Choose a coordinate 
system with the origin at Petoskey, the 
positive x direction to the east, and the 
positive y direction to the north. Let t = 0 at 
9:00 a.m. and θ be the angle between 
Robert’s velocity vector and the easterly 
direction and let ″M″ and ″R″ denote Mary 
and Robert, respectively. You can express 
the positions of Mary and Robert as 
functions of time and then equate their 
north (y) and east (x) coordinates at the 
time they rendezvous. 
 
 
Express Mary’s position as a 
function of time: 
 
( ) jjr ˆ8ˆMM ttv ==r 
where Mr
r
is in miles if t is in hours. 
 
Note that Robert’s initial position 
coordinates (xi, yi) are: 
 
(xi, yi) = (−13 mi, 22.5 mi) 
 
Express Robert’s position as a function of time: 
 
ji
jir
ˆ}]sin)1(6{5.22[ˆ}]cos)1(6{13[
ˆ)]1)(sin([ˆ1)]))(cos( [ RiRiR
θθ
θθ
−++−+−=
−++−+=
tt
tvytvxr
 
where Rr
r
is in miles if t is in hours. 
 
When Mary and Robert rendezvous, 
their coordinates will be the same. 
Equating their north and east 
coordinates yields: 
 
 East: –13 + 6t cosθ – 6 cosθ = 0 (1) 
 
 North: 22.5 + 6t sinθ – 6 sinθ = 8t (2) 
Solve equation (1) for cosθ : 
 ( )16
13cos −= tθ (3) 
 
Solve equation (2) for sinθ : 
 ( )16
5.228sin −
−=
t
tθ (4) 
Square and add equations (3) and (4) to obtain: 
 
( ) ( )
22
22
16
13
16
5.2281cossin ⎥⎦
⎤⎢⎣
⎡
−+⎥⎦
⎤⎢⎣
⎡
−
−==+
tt
tθθ 
 
Simplify to obtain a quadratic 
equation in t: 
 
063928828 2 =+− tt 
Solve (you could use your 
calculator’s ″solver″ function) this min15h3h24.3 ==t 
Motion in One and Two Dimensions 
 
 
153
equation for the smallest value of t 
(both roots are positive) to obtain: 
 
Now you can find the distance 
traveled due north by Mary: 
( )( ) mi9.25h24.3mi/h8MM === tvr 
 
Finally, solving equation (3) for θ and substituting 3.24 h for t yields: 
 
( ) ( ) °=⎥⎦
⎤⎢⎣
⎡
−=⎥⎦
⎤⎢⎣
⎡
−=
−− 7.14
124.36
13cos
16
13cos 11
t
θ 
and so Robert should head east. ofnorth 7.14 ° 
 
Remarks: Another solution that does not depend on the components of the vectors 
utilizes the law of cosines to find the time t at which Mary and Robert meet and then 
uses the law of sines to find the direction that Robert must head in order to 
rendezvous with Mary. 
 
Relative Velocity 
 
63 •• 
Picture the Problem Choose a coordinate 
system in which north is the positive y 
direction and east is the positive x 
direction. Let θ be the angle between 
north and the direction of the plane’s 
heading. The velocity of the plane relative 
to the ground, PGv
r
, is the sum of the 
velocity of the plane relative to the air, 
PAv
r
, and the velocity of the air relative to 
the ground, AGv
r
. i.e., 
 AGPAPG vvv
rrr += 
The pilot must head in such a direction that 
the east-west component of PGv
r
is zero in 
order to make the plane fly due north. 
 
 
 
 
(a) From the diagram one can see 
that: 
 
vAG cos 45° = vPA sinθ 
 
Solve for and evaluate θ : 
north of west 1.13
km/h250
km/h6.56sin 1
°=
⎟⎟⎠
⎞
⎜⎜⎝
⎛= −θ
 
 
(b) Because the plane is headed due 
north, add the north components of PG
vr = (250 km/h) cos 13.1° 
 + (80 km/h) sin 45° 
Chapter 3 
 
 
154 
AGPA and vv
rr
 to determine the 
plane’s ground speed: 
 = km/h300 
 
64 •• 
Picture the Problem Let SBv
r
represent the 
velocity of the swimmer relative to the 
bank; SWv
r
the velocity of the swimmer 
relative to the water; and WBv
r
the velocity 
of the water relative to the shore; i.e., 
 
SBv
r
= SWv
r
+ WBv
r
 
 
The current of the river causes the 
swimmer to drift downstream. 
 
(a) The triangles shown in the figure 
are similar right triangles. Set up a 
proportion between their sides and 
solve for the speed of the water 
relative to the bank: 
 
m80
m40
SW
WB =
v
v
 
and 
( ) m/s800.0m/s6.121WB ==v 
 
(b) Use the Pythagorean Theorem to 
solve for the swimmer’s speed 
relative to the shore: 
 
( ) ( )
m/s79.1
m/s8.0m/s6.1 22
2
WS
2
SWSB
=
+=
+= vvv
 
 
(c) The swimmer should head in a 
direction such that the upstream 
component of her velocity is equal 
to the speed of the water relative to 
the shore: 
 
Use a trigonometric function to 
evaluate θ: °=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 0.30
m/s1.6
m/s0.8sin 1θ 
 
Motion in One and Two Dimensions 
 
 
155
*65 •• 
 
Use the diagram to express the 
condition relating the eastward 
component of AGv
r
 and the 
westward component of .PAv
r
This 
must be satisfied if the plane is to 
stay on its northerly course. [Note: 
this is equivalent to equating the x-
components of equation (1).] 
 
(50 km/h) cos 45° = (240 km/h) sinθ 
 
Now solve for θ to obtain: 
 
( ) °=⎥⎦
⎤⎢⎣
⎡ °= − 47.8
km/h240
cos45km/h50sin 1θ 
 
Add the north components of PAv
r
 
and AGv
r
 to find the velocity of the 
plane relative to the ground: 
 
vPG + vAGsin45° = vPAcos8.47° 
and 
vPG = (240 km/h)cos 8.47° 
 − (50 km/h)sin 45° 
 = 202 km/h 
 
Finally, find the time of flight: 
h57.2
km/h202
km520
travelleddistance
PG
flight
==
=
v
t
 
 
Picture the Problem Let the velocity of 
the plane relative to the ground be 
represented by ;PGv
r
the velocity of the 
plane relative to the air by ,PAv
r
and the 
velocity of the air relative to the ground by 
.AGv
r
 Then 
AGPAPG vvv
rrr += (1) 
Choose a coordinate system with the origin 
at point A, the positive x direction to the 
east, and the positive y direction to the 
north. θ is the angle between north and the 
direction of the plane’s heading.The pilot 
must head so that the east-west component 
of PGv
r
is zero in order to make the plane fly 
due north. 
Chapter 3 
 
 
156 
66 •• 
Picture the Problem Let BSv
r
be the 
velocity of the boat relative to the shore; 
BWv
r
be the velocity of the boat relative to 
the water; and WSv
r
represent the velocity of 
the water relative to the shore. 
Independently of whether the boat is going 
upstream or downstream: 
BSv
r
= BWv
r
+ WSv
r
 
Going upstream, the speed of the boat 
relative to the shore is reduced by the speed 
of the water relative to the shore. 
 
 
Going downstream, the speed of the boat 
relative to the shore is increased by the 
same amount. 
 
 
For the upstream leg of the trip: 
 
vBS = vBW − vWS 
For the downstream leg of the trip: 
 
vBS = vBW + vWS 
 
Express the total time for the trip in 
terms of the times for its upstream 
and downstream legs: 
 WSBWWSBW
downstreamupstreamtotal
vv
L
vv
L
ttt
++−=
+=
 
 
Multiply both sides of the equation 
by (vBW − vWS)(vBW + vWS) (the 
product of the denominators) and 
rearrange the terms to obtain: 
 
 
02 2WSBW
total
2
BW =−− vvt
Lv 
Solve the quadratic equation for 
vBW. (Only the positive root is 
physically meaningful.) 
vBW = km/h18.5 
 
 
67 •• 
Picture the Problem Let pgv
r
be the 
velocity of the plane relative to the ground; 
agv
r
be the velocity of the air relative to the 
ground; and pav
r
the velocity of the plane 
relative to the air. Then, pgv
r
= pav
r
+ 
.agv
r
The wind will affect the flight times 
differently along these two paths. 
 
Motion in One and Two Dimensions 
 
 
157
The velocity of the plane, relative to 
the ground, on its eastbound leg is 
equal to its velocity on its 
westbound leg. Using the diagram, 
find the velocity of the plane 
relative to the ground for both 
directions: 
 
( ) ( ) m/s1.14m/s5m/s15 22
2
ag
2
papg
=−=
−= vvv
 
Express the time for the east-west 
roundtrip in terms of the distances 
and velocities for the two legs: 
 
s141
m/s14.1
m102
circle the of radius
circletheofradius
3
westboundpg,
eastboundpg,
westboundeastboundEWtrip,round
=×=
+
=
+=
v
v
ttt
 
 
Use the distances and velocities for the two legs to express and evaluate the time for 
the north-south roundtrip: 
 
s150
m/s) (5m/s) (15
m10
m/s) (5m/s) (15
m10
circletheofradiuscircletheofradius
33
southboundpg,northboundpg,
southboundnorthboundNStrip,round
=++−=
+=+=
vv
ttt
 
 wind. theacross planeyour fly shouldyou , Because NStrip,roundEWroundtrip, tt < 
 
 
68 • 
Picture the Problem This is a relative 
velocity problem. The given quantities are 
the direction of the velocity of the plane 
relative to the ground and the velocity 
(magnitude and direction) of the air relative 
to the ground. Asked for is the direction of 
the velocity of the air relative to the 
ground. Using ,AGPAPG vvv
rrr += draw a 
vector addition diagram and solve for the 
unknown quantity. 
 
 
Calculate the heading the pilot must 
take: °== − 5.11kts150
kts30sin 1θ 
Because this is also the angle of the 
plane's heading clockwise from 
north, it is also its azimuth or the 
required true heading: 
Az = (011.5°) 
Chapter 3 
 
 
158 
*69 •• 
Picture the Problem The position of B 
relative to A is the vector from A to B; i.e., 
ABAB rrr
rrr −= 
The velocity of B relative to A is 
dtd ABAB rv
rr = 
and the acceleration of B relative to A is 
dtd ABAB va
rr = 
Choose a coordinate system with the origin 
at the intersection, the positive x direction 
to the east, and the positive y direction to 
the north. 
 
 
 
(a) Find ABAB and,, rrr
rrr
: 
 
 
 
 
 
 
 
( )[ ]
( )[ ]ir
jr
ˆm/s20
ˆm/s2m40
A
22
2
1
B
t
t
=
−=
r
r
 
and 
( )[ ]
( )[ ]j
i
r
ˆm/s2m40
ˆm/s20
22
2
1
ABAB
t
t
rr
−+
−=
−= rrr
 
 
Evaluate ABr
r
at t = 6 s: 
 
jir ˆ m) (4 ˆ m) (120)s6(AB +=r 
(b) Find dtd ABAB rv
rr = : 
 
 
 
 
 
( ){ }[
( ){ } ]
ji
j
irv
ˆ)m/s 2(ˆm/s) 20(
ˆm/s 2m 40
m/s 20
2
22
2
1
AB
AB
t
t
t
dt
d
dt
d
−+−=
−+
−== r
rr
 
Evaluate ABv
r
at t = 6 s: 
 
( ) ( ) ( ) jiv ˆm/s12ˆm/s20s6AB −−=r 
(c) Find dtd ABAB va
rr = : 
 
 
 
 
[ ]
( )j
jia
ˆm/s2
ˆ )m/s 2( ˆ m/s) 20(
2
2
AB
−=
−+−= t
dt
dr
 
Note that ABa
r
is independent of time. 
 
*70 ••• 
Picture the Problem Let h and h′ represent the heights from which the ball is dropped 
and to which it rebounds, respectively. Let v and v′ represent the speeds with which the 
ball strikes the racket and rebounds from it. We can use a constant-acceleration equation 
to relate the pre- and post-collision speeds of the ball to its drop and rebound heights. 
 
Motion in One and Two Dimensions 
 
 
159
(a) Using a constant-acceleration 
equation, relate the impact speed of 
the ball to the distance it has fallen: 
 
ghvv 220
2 += 
or, because v0 = 0, 
ghv 2= 
 
Relate the rebound speed of the ball 
to the height to which it rebounds: 
 
gh'v'v 222 −= 
or because v = 0, 
gh'v' 2= 
 
Divide the second of these equations 
by the first to obtain: 
h
h'
gh
gh'
v
v' ==
2
2
 
 
Substitute for h′ and evaluate the 
ratio of the speeds: 
 
8.064.0 ==
h
h
v
v'
 ⇒ vv' 8.0= 
 
(b) Call the speed of the racket V. In a reference frame where the racket is 
unmoving, the ball initially has speed V, moving toward the racket. After it 
"bounces" from the racket, it will have speed 0.8 V, moving away from the racket. 
 
In the reference frame where the 
racket is moving and the ball 
initially unmoving, we need to add 
the speed of the racket to the speed 
of the ball in the racket's rest frame. 
Therefore, the ball's speed is: 
mi/h100
m/s548.18.0
≈
==+= VVVv'
 
This speed is close to that of a tennis pro’s 
serve. Note that this result tells us that the 
ball is moving significantly faster than the 
racket. 
 
(c) 
racket. theas
fast as e than twicmore movenever can ball the),(part in result theFrom b
 
 
Circular Motion and Centripetal Acceleration 
 
71 • 
Picture the Problem We can use the definition of centripetal acceleration to express ac in 
terms of the speed of the tip of the minute hand. We can find the tangential speed of the 
tip of the minute hand by using the distance it travels each revolution and the time it takes 
to complete each revolution. 
 
Express the acceleration of the tip of 
the minute hand of the clock as a 
function of the length of the hand 
and the speed of its tip: 
 
R
va
2
c = 
Use the distance the minute hand 
travels every hour to express its 
speed: 
 
T
Rv π2= 
Chapter 3 
 
 
160 
Substitute to obtain: 
2
2
c
4
T
Ra π= 
 
Substitute numerical values and 
evaluate ac: 
( )
( ) 262
2
c m/s1052.1s3600
m5.04 −×== πa 
 
Express the ratio of ac to g: 7
2
26
c 1055.1
m/s9.81
m/s101.52 −− ×=×=
g
a
 
 
72 • 
Picture the Problem The diagram shows 
the centripetal and tangential accelerations 
experienced by the test tube. The tangential 
acceleration will be zero when the 
centrifuge reaches its maximum speed. The 
centripetal acceleration increases as the 
tangential speed of the centrifuge increases. 
We can use the definition of centripetal 
acceleration to express ac in terms of the 
speed of the test tube. We can find the 
tangential speed of the test tube by using 
the distanceit travels each revolution and 
the time it takes to complete each 
revolution. The tangential acceleration can 
be found from the change in the tangential 
speed as the centrifuge is spinning up. 
 
 
 
 
 
(a) Express the acceleration of the 
centrifuge arm as a function of the 
length of its arm and the speed of 
the test tube: 
 
R
va
2
c = 
Use the distance the test tube travels 
every revolution to express its 
speed: 
 
T
Rv π2= 
Substitute to obtain: 
2
2
c
4
T
Ra π= 
 
Substitute numerical values and 
evaluate ac: 
( )
25
2
2
c
m/s1070.3
min
s60
rev15000
min1
m15.04
×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×
= πa
 
 
Motion in One and Two Dimensions 
 
 
161
(b) Express the tangential 
acceleration in terms of the 
difference between the final and 
initial tangential speeds: 
 
tT
R
t
T
R
t
vva ∆=∆
−
=∆
−= π
π
20
2
if
t 
 
Substitute numerical values and 
evaluate aT: 
( )
( )
2
t
m/s14.3
s75
min
s60
rev15000
min1
m15.02
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×
= πa
 
 
73 • 
Picture the Problem The diagram includes 
a pictorial representation of the earth in its 
orbit about the sun and a force diagram 
showing the force on an object at the 
equator that is due to the earth’s rotation, 
,RF
r
 and the force on the object due to the 
orbital motion of the earth about the sun, 
.oF
r
Because these are centripetal forces, 
we can calculate the accelerations they 
require from the speeds and radii associated 
with the two circular motions. 
 
 
 
 
Express the radial acceleration due 
to the rotation of the earth: R
va
2
R
R = 
Express the speed of the object on 
the equator in terms of the radius of 
the earth R and the period of the 
earth’s rotation TR: 
 
R
R
2
T
Rv π= 
Substitute for vR in the expression 
for aR to obtain: 2R
2
R
4
T
Ra π= 
 
Substitute numerical values and 
evaluate aR: 
( )
( )
g
a
3
22
2
32
R
1044.3
m/s1037.3
h1
s3600h24
m1063704
−
−
×=
×=
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
×= π
 
Note that this effect gives rise to the well-
known latitude correction for g. 
Chapter 3 
 
 
162 
Express the radial acceleration due 
to the orbital motion of the earth: r
va
2
o
o = 
 
Express the speed of the object on 
the equator in terms of the earth-sun 
distance r and the period of the 
earth’s motion about the sun To: 
 
o
o
2
T
rv π= 
Substitute for vo in the expression 
for ao to obtain: 2o
2
o
4
T
ra π= 
 
Substitute numerical values and evaluate 
ao: 
( )
( )
g
a
423
2
112
o
1007.6m/s1095.5
h1
s3600
d1
h24d365
m10.514
−− ×=×=
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛
×= π
 
 
74 •• 
Picture the Problem We can relate the acceleration of the moon toward the earth to its 
orbital speed and distance from the earth. Its orbital speed can be expressed in terms of its 
distance from the earth and its orbital period. From tables of astronomical data, we find 
that the sidereal period of the moon is 27.3 d and that its mean distance from the earth is 
3.84×108 m. 
 
Express the centripetal acceleration 
of the moon: 
 r
va
2
c = 
Express the orbital speed of the 
moon: 
 T
rv π2= 
 
Substitute to obtain: 
2
2
c
4
T
ra π= 
 
Substitute numerical values and 
evaluate ac: 
( )
g
πa
4
23
2
82
c
1078.2
m/s1072.2
h
s3600
d
h24d27.3
m103.844
−
−
×=
×=
⎟⎠
⎞⎜⎝
⎛ ××
×=
 
 
Motion in One and Two Dimensions 
 
 
163
Remarks: Note that 
moon to earth fromdistance 
earthofradius
g
ac = (ac is just the acceleration 
due to the earth’s gravity evaluated at the moon’s position). This is Newton’s 
famous ″falling apple″ observation. 
 
75 • 
Picture the Problem We can find the number of revolutions the ball makes in a given 
period of time from its speed and the radius of the circle along which it moves. Because 
the ball’s centripetal acceleration is related to its speed, we can use this relationship to 
express its speed. 
 
Express the number of revolutions 
per minute made by the ball in terms 
of the circumference c of the circle 
and the distance x the ball travels in 
time t: 
 
c
xn = (1) 
Relate the centripetal acceleration of 
the ball to its speed and the radius of 
its circular path: 
 
R
vga
2
c == 
Solve for the speed of the ball: 
 
Rgv = 
 
Express the distance x traveled in 
time t at speed v: 
 
vtx = 
 
Substitute to obtain: 
 
tRgx = 
 
The distance traveled per revolution 
is the circumference c of the circle: 
Rc π2= 
 
 
Substitute in equation (1) to obtain: 
t
R
g
R
tRgn ππ 2
1
2
== 
 
Substitute numerical values and 
evaluate n: ( ) 12 min33.4s60
m0.8
m/s9.81
2
1 −== πn
Remarks: The ball will oscillate at the end of this string as a simple pendulum with 
a period equal to 1/n. 
 
Projectile Motion and Projectile Range 
 
76 • 
Picture the Problem Neglecting air resistance, the accelerations of the ball are constant 
and the horizontal and vertical motions of the ball are independent of each other. We can 
use the horizontal motion to determine the time-of-flight and then use this information to 
determine the distance the ball drops. Choose a coordinate system in which the origin is 
at the point of release of the ball, downward is the positive y direction, and the horizontal 
Chapter 3 
 
 
164 
direction is the positive x direction. 
 
Express the vertical displacement of 
the ball: 
( )2210 tatvy yy ∆+∆=∆ 
or, because v0y = 0 and ay = g, ( )221 tgy ∆=∆ 
 
Find the time of flight from 
vx = ∆x/∆t: 
 ( )( )
( )( ) s0.473m/km1000km/h140
s/h3600m18.4 ==
∆=∆
xv
xt
 
 
Substitute to find the vertical 
displacement in 0.473 s: 
( )( ) m1.10s0.473m/s9.81 2221 ==∆y 
 
77 • 
Picture the Problem In the absence of air resistance, the maximum height achieved by a 
projectile depends on the vertical component of its initial velocity. 
 
The vertical component of the 
projectile’s initial velocity is: 
v0y = v0 sinθ0 
 
 
Use the constant-acceleration 
equation: 
 
 v y
2 = v0 y2 + 2ay∆y 
Set vy = 0, a = −g, and ∆y = h to 
obtain: 
 
( )
g
vh
2
sin 200 θ= 
 
*78 •• 
Picture the Problem Choose the 
coordinate system shown to the right. 
Because, in the absence of air resistance, 
the horizontal and vertical speeds are 
independent of each other, we can use 
constant-acceleration equations to relate 
the impact speed of the projectile to its 
components. 
 
 
 
The horizontal and vertical velocity 
components are: 
 
 
v0x = vx= v0cosθ 
and 
v0y = v0sinθ 
Using a constant-acceleration 
equation, relate the vertical 
yavv yyy ∆+= 2202 
or, because ay = −g and ∆y = −h, 
Motion in One and Two Dimensions 
 
 
165
component of the velocity to the 
vertical displacement of the 
projectile: 
 
( ) ghvvy 2sin 202 += θ 
Express the relationship between the 
magnitude of a velocity vector and 
its components, substitute for the 
components, and simplify to obtain: 
 
 
( )
( )
ghv
ghv
vvvvv yyx
2
2cossin
cos
2
0
222
0
22
0
222
+=
++=
+=+=
θθ
θ
 
 
Substitute for v: ( ) ghvv 22.1 2020 += 
 
Set v = 1.2 v0, h = 40 m and solve 
for v0: 
m/s2.420 =v 
 
Remarks: Note that v is independent of θ. This will be more obvious once 
conservation of energy has been studied. 
 
79 •• 
Picture the Problem Example 3-12 shows that the dart willhit the monkey unless the 
dart hits the ground before reaching the monkey’s line of fall. What initial speed does the 
dart need in order to just reach the monkey’s line of fall? First, we will calculate the fall 
time of the monkey, and then we will calculate the horizontal component of the dart’s 
velocity. 
 
Using a constant-acceleration 
equation, relate the monkey’s fall 
distance to the fall time: 
 
2
2
1 gth = 
 
Solve for the time for 
the monkey to fall to the ground: 
 g
ht 2= 
 
Substitute numerical values and 
evaluate t: 
( ) s51.1
m/s9.81
m2.112
2 ==t 
 
Let θ be the angle the barrel of the 
dart gun makes with the horizontal. 
Then: 
 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 3.11
m50
m10tan 1θ 
Use the fact that the horizontal 
velocity is constant to determine v0: 
( ) m/s8.33
cos11.3
s1.51m50 
cos0
=°== θ
xvv 
 
Chapter 3 
 
 
166 
80 •• 
Picture the Problem Choose the 
coordinate system shown in the figure to 
the right. In the absence of air resistance, 
the projectile experiences constant 
acceleration in both the x and y directions. 
We can use the constant-acceleration 
equations to express the x and y 
coordinates of the projectile along its 
trajectory as functions of time. The 
elimination of the parameter t will yield an 
expression for y as a function of x that we 
can evaluate at (R, 0) and (R/2, h). Solving 
these equations simultaneously will yield 
an expression for θ. 
 
 
 
Express the position coordinates 
of the projectile along its flight 
path in terms of the parameter t: 
 
( )tvx θcos0= 
and ( ) 2210 sin gttvy −= θ 
 
Eliminate the parameter t to 
obtain: 
 
( ) 222
0 cos2
tan x
v
gxy θθ −= (1) 
 
Evaluate equation (1) at (R, 0) to 
obtain: 
 g
vR θθ cossin2
2
0= 
 
Evaluate equation (1) at (R/2, h) 
to obtain: 
 
( )
g
vh
2
sin 20 θ= 
 
Equate R and h and solve the 
resulting equation for θ : 
 
( ) °== − 0.764tan 1θ 
 
Remarks: Note that this result is independent of v0. 
 
81 •• 
Picture the Problem In the absence of air 
resistance, the motion of the ball is 
uniformly accelerated and its horizontal 
and vertical motions are independent of 
each other. Choose the coordinate system 
shown in the figure to the right and use 
constant-acceleration equations to relate the 
x and y components of the ball’s initial 
velocity. 
 
Use the components of v0 to express 
θ in terms of v0x and v0y: 
 x
y
v
v
0
01tan−=θ (1) 
Motion in One and Two Dimensions 
 
 
167
Use the Pythagorean relationship 
between the velocity and its 
components to express v0: 
 
2
0
2
00 yx vvv += (2) 
 
Using a constant-acceleration 
equation, express the vertical speed 
of the projectile as a function of its 
initial upward speed and time into 
the flight: 
 
vy= v0y+ ay t 
 
Because vy = 0 halfway through the 
flight (at maximum elevation): 
 
v0y = (9.81 m/s2)(1.22 s) = 12.0 m/s 
 
Determine v0x: 
 m/s4.16s2.44
m40
0x ==∆
∆=
t
xv 
 
Substitute in equation (2) and 
evaluate v0: 
 
( ) ( )
m/s3.20
m/s0.12m/s4.16 220
=
+=v
 
 
Substitute in equation (1) and 
evaluate θ : 
 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 2.36
m/s16.4
m/s12.0tan 1θ 
 
*82 •• 
Picture the Problem In the absence of 
friction, the acceleration of the ball is 
constant and we can use the constant- 
acceleration equations to describe its 
motion. The figure shows the launch 
conditions and an appropriate coordinate 
system. The speeds v, vx, and vy are related 
through the Pythagorean Theorem. 
 
 
The squares of the vertical and 
horizontal components of the 
object’s velocity are: 
 θ
θ
22
0
2
22
0
2
cos
and
2sin
vv
ghvv
x
y
=
−=
 
 
The relationship between these 
variables is: 
 
222
yx vvv += 
Substitute and simplify to obtain: ghvv 220
2 −= 
Note that v is independent of θ ... as was 
to be shown. 
 
Chapter 3 
 
 
168 
83 •• 
Picture the Problem In the absence of air 
resistance, the projectile experiences 
constant acceleration during its flight and 
we can use constant-acceleration equations 
to relate the speeds at half the maximum 
height and at the maximum height to the 
launch angle θ of the projectile. 
 
 
 
The angle the initial velocity makes 
with the horizontal is related to the 
initial velocity components. x
y
v
v
0
0tan =θ 
Write the equation 
y,avv yy ∆+= 2202 for ∆y = h and 
vy = 0: 
 
 20 20 ghvhy y −=⇒=∆ (1) 
Write the equation 
y,avv yy ∆+= 2202 for ∆y = h/2: 
 
 
2
2 
2
2
0
2 hgvvhy yy −=⇒=∆ (2) 
We are given vy = (3/4)v0. Square 
both sides and express this using the 
components of the velocity. The x 
component of the velocity remains 
constant. 
 
( ) 
4
3 2
0
2
0
2
22
0 yxyx vvvv +⎟⎠
⎞⎜⎝
⎛=+ (3) 
where we have used vx = v0x . 
(Equations 1, 2, and 3 constitute three equations and four unknowns v0x, v0y, vy, and h. To 
solve for any of these unknowns, we first need a fourth equation. However, to solve for 
the ratio (v0y/v0x) of two of the unknowns, the three equations are sufficient. That is 
because dividing both sides of each equation by v0x
2 gives three equations and three 
unknowns vy/v0x, v0y/v0x, and h/ .2x0v 
 
Solve equation 2 for gh and 
substitute in equation 1: ( )22020 2 hyy vvv −= ⇒ 2
2
02 y
y
v
v = 
Substitute for vy
2 in equation 3: ( )202022020 4321 yxyx vvvv +⎟⎠⎞⎜⎝⎛=+ 
Motion in One and Two Dimensions 
 
 
169
Divide both sides by v0x
2 and solve 
for v0y/v0x to obtain: 
 
 
 
7 
and
 1
16
9
2
11
0
0
2
0
2
0
2
0
2
0
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=+
x
y
x
y
x
y
v
v
v
v
v
v
 
Using tan θ = v0y/v0x, solve for θ : ( ) °==⎟⎟⎠⎞⎜⎜⎝⎛= −− 3.697tantan 1001 xyvvθ 
 
84 • 
Picture the Problem The horizontal speed 
of the crate, in the absence of air resistance, 
is constant and equal to the speed of the 
cargo plane. Choose a coordinate system in 
which the direction the plane is moving is 
the positive x direction and downward is 
the positive y direction and apply the 
constant-acceleration equations to describe 
the crate’s displacements at any time 
during its flight. 
 
 
(a) Using a constant-acceleration 
equation, relate the vertical 
displacement of the crate ∆y to the 
time of fall ∆t: 
 
( )2210 tgtvy y ∆+∆=∆ 
or, because v0y = 0, ( )221 tgy ∆=∆ 
 
Solve for ∆t: 
g
yt ∆=∆ 2 
 
Substitute numerical values and 
evaluate ∆t: ( ) s5.49
m/s81.9
m10122
2
3
=×=∆t 
 
(b) The horizontal distance traveled 
in 49.5 s is: 
 
 
( ) ( )
km4.12
s5.49
s3600
h1km/h900
0
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
∆=∆= tvxR x
 
 
(c) Because the velocity of the plane 
is constant, it will be directly over 
the crate when it hits the ground; 
i.e., the distance to the aircraft will 
be the elevation of the aircraft. 
km0.12=∆y 
Chapter 3 
 
 
170 
*85 •• 
Picture the Problem In the absence of air 
resistance, the accelerations of both Wiley 
Coyote and the Roadrunner are constant 
and we can use constant-acceleration 
equations to express their coordinates at 
any time during their leaps across the 
gorge. By eliminating the parameter t 
between these equations, we can obtain an 
expression that relates their y coordinates 
to their x coordinates and that we can solve 
for their launch angles. 
 
 
(a) Using constant-acceleration 
equations, express the x coordinate 
of the Roadrunner while it is in 
flight across the gorge: 
 
2
2
1
00 tatvxx xx ++= 
or, because x0 = 0, ax =0 and 
v0x = v0 cosθ0, ( )tvx 00 cosθ= 
 
Using constant-acceleration 
equations, express the y coordinate 
of the Roadrunner while it is in 
flight across the gorge: 
 
2
2
1
00 tatvyy yy ++= 
or, because y0 = 0, ay =−g and 
v0y = v0 sinθ0, ( ) 22100 sin gttvy −= θ 
 
Eliminate the parameter t to obtain: 
 ( ) 2
0
22
0
0 cos2
tan x
v
gxy θθ −= (1) 
 
Letting R represent the 
Roadrunner’s range and using the 
trigonometric identity 
sin2θ = 2sinθ cosθ, solve for and 
evaluate its launch speed: 
( )( )
m/s0.18
30sin
m/s9.81m5.16
2sin
2
0
0
=
°== θ
Rgv
 
 
(b) Letting R represent Wiley’s 
range, solve equation (1) for his 
launch angle: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 2
0
1
0 sin2
1
v
Rgθ 
 
Substitute numerical values and 
evaluate θ0: ( )( )( )
°=
⎥⎦
⎤⎢⎣
⎡= −
0.13
m/s.081
m/s9.81m14.5sin
2
1
2
2
1
0θ
 
 
Motion in One and Two Dimensions 
 
 
171
86 • 
Picture the Problem Because, in the 
absence of air resistance, the vertical and 
horizontal accelerations of the cannonball 
are constant, we can use constant-
acceleration equations to express the ball’s 
position and velocity as functions of time 
and acceleration. The maximum height of 
the ball and its time-of-flight are related to 
the components of its launch velocity. 
 
 
 
(a) Using a constant-acceleration 
equation, relate h to the initial and 
final speeds of the cannonball: 
 
yavv yy ∆+= 2202 
or, because v = 0 and ay = −g, 
ygv y ∆−= 20 20 
 
 Find the vertical component of the 
firing speed: 
 
 
v0y = v0sinθ = (300 m/s)sin 45° 
 = 212 m/s 
Solve for and evaluate h: ( )( ) km29.2m/s81.92 m/s2122 2
22
0 ===
g
v
h y 
 
(b) The total flight time is: 
 
 ( ) s2.43
m/s9.81
m/s21222
2
2
0
updnup
===
=+=∆
g
v
tttt
y 
 
(c) Express the x coordinate of the 
ball as a function of time: 
 
( ) tvtvx x ∆=∆= θcos00 
 
Evaluate x (= R) when ∆t = 43.2 s: ( )[ ]( )
km9.16
s43.2cos45m/s300
=
°=x
 
 
87 •• 
Picture the Problem Choose a coordinate 
system in which the origin is at the base of 
the tower and the x- and y-axes are as 
shown in the figure to the right. In the 
absence of air resistance, the horizontal 
speed of the stone will remain constant 
during its fall and a constant-acceleration 
equation can be used to determine the time 
of fall. The final velocity of the stone will 
be the vector sum of its x and y 
components. 
 
 
Chapter 3 
 
 
172 
(a) Using a constant-acceleration 
equation, express the vertical 
displacement of the stone (the 
height of the tower) as a function of 
the fall time: 
 
 
( )2210 tatvy yy ∆+∆=∆ 
or, because v0y = 0 and a = −g, ( )221 tgy ∆−=∆ 
Solve for and evaluate the time of 
fall: 
 
( ) s21.2
m/s81.9
m2422
2 =−−=∆−=∆ g
yt 
 
Use the definition of average 
velocity to find the velocity with 
which the stone was thrown from 
the tower: 
 
s/m14.8
s2.21
m18
0 ==∆
∆≡=
t
xvv xx 
 
(b) Find the y component of the 
stone’s velocity after 2.21 s: 
 
 
 
m/s 21.7
 s) m/s2)(2.21 (9.81 0
0
−=
−=
−= gtvv yy
 
Express v in terms of its 
components: 
22
yx vvv += 
 
Substitute numerical values and 
evaluate v: ( ) ( )
m/s2.23
m/s7.21m/s14.8 22
=
−+=v
 
 
88 •• 
Picture the Problem In the absence of air resistance, the acceleration of the projectile is 
constant and its horizontal and vertical motions are independent of each other. We can 
use constant-acceleration equations to express the horizontal and vertical displacements 
of the projectile in terms of its time-of-flight. 
 
Using a constant-acceleration 
equation, express the horizontal 
displacement of the projectile as a 
function of time: 
 
( )2210 tatvx xx ∆+∆=∆ 
or, because v0x = v0cosθ and ax = 0, ( ) tvx ∆=∆ θcos0 
 
Using a constant-acceleration 
equation, express the vertical 
displacement of the projectile as a 
function of time: 
 
 
( )2210 tatvy yy ∆+∆=∆ 
or, because v0y = v0sinθ and ay = −g, ( ) ( )2210 cos tgtvy ∆−∆=∆ θ 
Substitute numerical values to 
obtain the quadratic equation: 
 
( )( )
( )( )2221 m/s81.9
60sinm/s60m200
t
t
∆−
∆°=−
 
 
Solve for ∆t: 
 
∆t = 13.6 s 
 
Motion in One and Two Dimensions 
 
 
173
Substitute for ∆t and evaluate the 
horizontal distance traveled by the 
projectile: 
∆x = (60 m/s)(cos60°)(13.6 s) 
 = m408 
 
89 •• 
Picture the Problem In the absence of air 
resistance, the acceleration of the 
cannonball is constant and its horizontal 
and vertical motions are independent of 
each other. Choose the origin of the 
coordinate system to be at the base of the 
cliff and the axes directed as shown and 
use constant- acceleration equations to 
describe both the horizontal and vertical 
displacements of the cannonball. 
 
 
Express the direction of the velocity 
vector when the projectile strikes 
the ground: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
x
y
v
v1tanθ 
Express the vertical displacement 
using a constant-acceleration 
equation: 
 
 
 
( )2210 tatvy yy ∆+∆=∆ 
or, because v0y = 0 and ay = −g, ( )221 tgy ∆−=∆ 
 
Set ∆x = −∆y (R = −h) to obtain: ( )221 tgtvx x ∆=∆=∆ 
 
Solve for vx: 
 tgt
xvx ∆=∆
∆= 21 
 
Find the y component of the 
projectile as it hits the ground: 
 
xyy vtgtavv 20 −=∆−=∆+= 
 
Substitute and evaluate θ : ( ) °−=−=⎟⎟⎠
⎞
⎜⎜⎝
⎛= −− 4.632tantan 11
x
y
v
vθ 
 
90 • 
Picture the Problem In the absence of air 
resistance, the vertical and horizontal 
motions of the projectile experience 
constant accelerations and are independent 
of each other. Use a coordinate system in 
which up is the positive y direction and 
horizontal is the positive x direction and 
use constant-acceleration equations to 
describe the horizontal and vertical 
displacements of the projectile as functions 
of the time into the flight. 
 
 
 
Chapter 3 
 
 
174 
(a) Use a constant-acceleration 
equation to express the horizontal 
displacement of the projectile as a 
function of time: 
 
( ) tv
tvx x
∆=
∆=∆
θcos0
0 
 
Evaluate this expression when 
∆t = 6 s: 
 
( )( )( ) m900s6cos60m/s300 =°=∆x 
 
(b) Use a constant-acceleration 
equation to express the vertical 
displacement of the projectile as a 
function of time: 
 
( ) ( )2210 sin tgtvy ∆−∆=∆ θ 
 
Evaluate this expression when ∆t = 6 s: 
 
( )( )( ) ( )( ) km38.1s6m/s9.81s6sin60m/s300 2221 =−°=∆y 
 
91 •• 
Picture the Problem In the absence of air 
resistance, the acceleration of the projectile 
is constant and the horizontal and vertical 
motions are independent of each other. 
Choose the coordinate system shown in the 
figure with the origin at the base of the cliff 
and the axes oriented as shown and use 
constant-acceleration equations to find the 
range of the cannonball. 
 
Using a constant-acceleration 
equation, express the horizontal 
displacement of the cannonball as a 
function of time: 
 
( )2210 tatvx xx ∆+∆=∆ 
or, because v0x = v0cosθ and ax = 0, ( ) tvx ∆=∆ θcos0 
 
Using a constant-acceleration 
equation, express the vertical 
displacement of the cannonball as a 
function of time: 
 
 
 
( )2210 tatvy yy ∆+∆=∆ 
or, because y = −40 m, a = −g, and 
v0y = v0sinθ, ( )( )
( )( )2221 m/s81.9
30sinm/s42.2m04
t
t
∆−
∆°=−
 
 
Solve the quadratic equation for ∆t: 
 
∆t = 5.73 s 
Calculate the range: ( )( )( )
m209
s5.73cos30m/s42.2
=
°=∆= xR
 
 
Motion in One and Two Dimensions 
 
 
175
*92 •• 
Picture the Problem Choose a coordinate 
system in which the origin is at ground 
level. Let the positive